Page 153 - Class 10th Physics Chapter Electricity NCERT Book Questions and Answers

Click here for Hindi medium

Chapter : Electricity

...

Page : 200 

Question 1: 

What does an electric circuit mean?

Answer :

An electric circuit is the pathway in which current can flow. It consists of electric 

devices, switching devices, source of electricity, etc. that are connected by 

conducting wires.

Question 2: 

Define the unit of current.

Answer :

The unit of electric current is ampere (A). When 1 C of charge flows through a 

conductor in 1sec, it called 1 ampere (A) current.

𝐼 = 𝑄/𝑡

Question 3: 

Calculate the number of electrons constituting one coulomb of charge.

Answer :

We know that one electron possesses a charge of 1.6 × 10^−19 C.

𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 = 𝑇𝑜𝑡𝑎𝑙 𝑐ℎ𝑎𝑟𝑔𝑒 / 𝐶ℎ𝑎𝑟𝑔𝑒 𝑜𝑛 1 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛

= 1 / 1.6 × 10^−19
= 6.25 × 10^18

So, the number of electrons constituting one coulomb of charge is 6 × 10^18 .

Page : 202  

Question 1: 

Name a device that helps to maintain a potential difference across a conductor.

Answer :

A cell, battery, power supply, etc. helps to maintain a potential difference across 

a conductor.

Question 2: 

What is meant by saying that the potential difference between two points is 1 V?

Answer :

When 1 J of work is required to move a charge of 1 C from one point to another,

then it is said that the potential difference between the two points is 1 V.

𝑉 = 𝑊 / 𝑄

1 𝑉 = 1𝐽 / 1𝐶

Question 3: 

How much energy is given to each coulomb of charge passing through a 6 V 

battery?

Answer :

𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑊𝑜𝑟𝑘 𝑑𝑜𝑛𝑒 / 𝐶ℎ𝑎𝑟𝑔𝑒

or Work done (or Energy) = Potential Difference × Charge

So, Work done = 6 Volt × 1 Coulomb = 6 Joules
Page : 209  

Question 1: 

On what factors does the resistance of a conductor depend?

Answer :

The resistance of a conductor depends upon the following factors:

(i) Length of the conductor

(ii) Cross-sectional area of the conductor

(iii) Nature of the conductor

(iv) Temperature of the conductor.

Question 2: 

Will current flow more easily through a thick wire or a thin wire of the same 

material, when connected to the same source? Why?

Answer :

Resistance (R) is inversely proportional to the area of cross-section (A) of the 

wire. So, thicker the wire, lower is the resistance of the wire and vice-versa. 

Therefore, current can flow more easily through a thick wire than a thin wire.

𝑅 = 𝜌𝐿 / 𝐴

𝑅 ∝1𝐴

Question 3: 

Let the resistance of an electrical component remains constant while the potential 

difference across the two ends of the component decreases to half of its former 

value. What change will occur in the current through it?

Answer :

According to the Ohm’s law V = IR

If the resistance remains constant, V is directly proportional to I.

𝑉 ∝ 𝐼

Now, if potential difference is reduced to half of its value, the current also become 

half of its original value.

Question 4: 

Why are coils of electric toasters and electric irons made of an alloy rather than a 

pure metal?

Answer :

The resistivity of an alloy is higher than the pure metal and it does not corrode 

easily. Moreover, even at high temperatures, the alloys do not melt readily. 

Hence, the coils of heating appliances such as electric toasters and electric irons 

are made of an alloy rather than a pure metal.

Question 5: 

Use the data in Table 12.2 to answer the following –

(a) Which among iron and mercury is a better conductor?

(b) Which material is the best conductor?

Answer :

(a). Resistivity of iron (10.0 × 10–8 Ω 𝑚) is lesser than that of the mercury (94.0 

× 10–8 Ω 𝑚). So, iron is good conductor as compared to mercury.

(b). Silver has lowest resistivity, so it is the best conductor.
Page : 213  

Question 1: 

Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 

V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all 

connected in series.

Answer :

The required schematic diagram is given below:

Question 2: 

Redraw the circuit of Question 1, putting in an ammeter to measure the current 

through the resistors and a voltmeter to measure the potential difference across 

the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?

Answer :

Resisters are connected in series.

So, the net resistance in the circuit = 5 Ω + 8 Ω + 12 Ω = 25 Ω

Net potential = 6 V

Using Ohm’s law V = IR

                           ⟹ 6 = I × 25
                ⟹ 𝐼 = 6 / 25

                           ⟹ 𝐼 = 0.24 𝐴𝑚𝑝𝑒𝑟𝑒

Now for the 12 Ω resistor, current = 0.24 A

So, using Ohm’s law V = 0.24 × 12 V = 2.88 V

Hence, the reading in the ammeter is 0.24 and voltmeter is 2.88.

Page : 216  

Question 1: 

Judge the equivalent resistance when the following are connected in parallel –

(a) 1 Ω and 10^6 Ω, (b) 1 Ω and 10^3 Ω, and 10^6 Ω.

Answer :

(a). The net resistance in parallel is given by

Here, 𝑅1 = 1 Ω and 𝑅2 = 10^6 Ω

So,

(b). The net resistance in parallel is given by

Here, 𝑅1 = 1 Ω, 𝑅2 = 10^3 Ω and 𝑅3 = 10^6 Ω

So,

Question 2: 

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of 

resistance 500 Ω are connected in parallel to a 220 V source. What is the 

resistance of an electric iron connected to the same source that takes as much 

current as all three appliances, and what is the current through it?

Answer :

Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter

of resistance 500 Ω are connected in parallel.

The net resistance in parallel is given by

Here, 𝑅1 = 100 Ω, 𝑅2 = 50 Ω and 𝑅3 = 500 Ω

So,

Now, using Ohm’s law V = IR, we have

Hence, the resistance of electric iron is 31.25 Ω and current through it is 7.04 A.

Question 3: 

What are the advantages of connecting electrical devices in parallel with the 

battery instead of connecting them in series?

Answer :

In parallel there is no division of voltage among the appliances. The potential 

difference across each appliance is equal to the supplied voltage and the total 

effective resistance of the circuit can be reduced by connecting electrical

appliances in parallel.

Question 4: 

How can three resistors of resistances 2 Ω, 3 Ω, and 6 Ω be connected to give a 

total resistance of (a) 4 Ω, (b) 1 Ω?

Answer 4:

(a). To get total resistance 4 Ω, connect 3 Ω and 6 Ω resistors in parallel and 2 Ω

resistance in series with the resultant.

Since, 3 Ω and 6 Ω resistors in parallel, so the net resistance

or , R12 = 2 ohm
Now, the resultant 𝑅12 and 2 Ω resistors are in series. So the net resistance

𝑅 = 𝑅12 + 2 Ω = 2 + 2 = 4 Ω

(b). To get total resistance 1 Ω, connect 2 Ω, 3 Ω and 6 Ω resistors in parallel.

The net resistance in parallel is given by

Here, 𝑅1 = 2 Ω, 𝑅2 = 3 Ω and 𝑅3 = 6 Ω

So,

1/𝑅 =1/ 2 + 1/3 + 1/6

or , 1 / R = (3 + 2 + 1) / 6

or , 1 / R = 6 / 6

or , 𝑅 = 1 Ω

Question 5: 

What is (a) the highest, (b) the lowest total resistance that can be secured by 

combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

Answer :

Connecting resistors in series always gives maximum resistance and parallel 

gives minimum resistance.

(a). The highest total resistance is given by

𝑅 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω

(b). The lowest total resistance is given by

Page : 218 

Question 1: 

Why does the cord of an electric heater not glow while the heating element does?

Answer :

The heating element of an electric heater is a resistor. According to Joule’s law 

of heating, the amount of heat produced by it is proportional to its resistance.

H = I^2Rt

The resistance of the element of an electric heater is very high. As current flows 

through the heating element, it becomes too hot and glows red. On the other hand, 

the resistance of the cord is low. It does not become red when current flows 

through it.

Question 2: 

Compute the heat generated while transferring 96000 coulomb of charge in one 

hour through a potential difference of 50 V.

Answer :

According to Joule’s law of heating, the amount of heat produced is given by

H = VIt

Where, 

V = 50 V

𝐼 = 𝐶ℎ𝑎𝑟𝑔𝑒 / 𝑡𝑖𝑚𝑒
= 9600 𝑐𝑜𝑢𝑙𝑜𝑚𝑏 / 1 ℎ𝑟

= 9600 / 60 × 60
= 80 / 3 𝐴

and t = 1 hour = 60 × 60 seconds

So, 

𝐻 = 50 × 80 / 3 × 60 × 60

= 4800000 𝐽 = 4.8 × 10^6𝐽

Question 3 

An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat 

developed in 30 s.

Answer :

According to Joule’s law of heating, the amount of heat produced is given by

H = VIt

Where, 

V = IR = 5A × 20 Ω = 100 V

I = 5 A

and t = 30 seconds

So, 

𝐻 = 100 × 5 × 30 𝐽

= 15000 𝐽 = 1.5 × 10^4𝐽

Page : 220 

Question 1: 

What determines the rate at which energy is delivered by a current?

Answer :

The rate of consumption of electric energy in an electric appliance is called 

electric power. Hence, the rate at which energy is delivered by a current is the 

power of the appliance.

Question 2: 

An electric motor takes 5 A from a 220 V line. Determine the power of the motor 

and the energy consumed in 2 h.

Answer :

Power of the electric motor is given by

P = VI

Where, V = 220 V and I = 5 A

So, Power P = 220 × 5 = 1100 W

Now, the energy consumed = Power × time

Where, P = 1100 W

t = 2 hours = 2 × 60 × 60 seconds = 7200 seconds

So, the energy consumed E = 1100 × 7200 J = 7920000 J

Exercise

Question 1: 

A piece of wire of resistance R is cut into five equal parts. These parts are then

connected in parallel. If the equivalent resistance of this combination is R′, then 

the ratio R/R′ is –

(a) 1/25      (b) 1/5      (c) 5      (d) 25

Answer :

Resistance of a piece of wire is directly proportional to its length. If the piece of 

wire has a resistance R and the wire is cut into five equal parts.

The resistance of each part = R/5

All the five parts are connected in parallel. Hence, equivalent resistance (R′) is 

given as

Hence , the  option "d" इस correct .

Question 2: 

Which of the following terms does not represent electrical power in a circuit?

(a) I^2R      (b) IR^2      (c) VI        (d) V^2/R

Answer :

We know that electric power is given by P = VI … (i)

So, the option (c) is correct.

According to Ohm’s law, V = IR … (ii)

Now putting the value of V from (ii) in (i), we get

Power P = (IR) × I = I^2R

So, the option (a) is correct.

Now putting the value of I from (ii) in (i), we get

Power P = V(V/R) = V^2/R

So, the option (d) is correct.

Hence, the option (b) does not represent electrical power in a circuit.

Question 3: 

An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the

power consumed will be –

(a) 100 W   (b) 75 W       (c) 50 W       (d) 25 W

Answer :

Energy consumed by bulb ( 𝑃 ) =𝑉^2/R

⟹ 𝑅 =𝑉^2/P

Here, V = 220 V and P = 100 W

𝑅 =(220)^2/100 = 484 Ω

The resistance of the bulb remains constant if the supply voltage is reduced to 

110 V. If the bulb is operated on 110 V, then the energy consumed by it is given 

by the expression for power 

𝑃 =𝑉^2/R

=(110)^2/484
 =12100/484
 = 25 𝑊

Hence, the option (d) is correct.

Question 4: 

Two conducting wires of the same material and of equal lengths and equal 

diameters are first connected in series and then parallel in a circuit across the same 

potential difference. The ratio of heat produced in series and parallel 

combinations would be –

(a) 1:2      (b) 2:1       (c) 1:4          (d) 4:1

Answer :

Heat produced in the circuit is inversely proportional to the resistance R. Let RS

and RP be the equivalent resistances of the wires if connected in series and parallel

respectively. Let R be the resistance of each wire.

If the resistors are connected in parallel, the net resistance is given by

1/𝑅𝑃 = 1/𝑅 + 1/𝑅

⟹1/𝑅𝑃 = 2/𝑅

⟹ 𝑅𝑃 =𝑅/2

If the resistors are connected in series, the net resistance is given by

𝑅𝑠 = 𝑅 + 𝑅 = 2𝑅

Hence, for same potential difference V, the ratio of heat produced in the circuit 

is given by

Therefore, the ratio of heat produced in series and parallel combinations is 1:4.

Hence, the option (c) is correct.

Question 5: 

How is a voltmeter connected in the circuit to measure the potential difference

between two points?

Answer :

To measure the potential difference, a voltmeter should be connected in parallel.

Question 6: 

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will 

be the length of this wire to make its resistance 10 Ω? How much does the 

resistance change if the diameter is doubled?

Answer :

Resistance (R) of a copper wire of length l and cross-section A is given by the

expression,

Hence , the new resistance wiil become 1/4 times th  original resistance .

Question 7: 

The values of current I flowing in a given resistor for the corresponding values of

potential difference V across the resistor are given below –

I (amperes) 0.5 1.0 2.0 3.0 4.0

V (volts) 1.6 3.4 6.7 10.2 13.2

Plot a graph between V and I and calculate the resistance of that resistor.

Answer :

The plot between voltage and current is called VI characteristic. The voltage is 

plotted on x-axis and current is plotted on y-axis.

The slope of the line gives the value of resistance (R)

𝑠𝑙𝑜𝑝𝑒 =1/𝑅 =𝐵𝐶/𝐴𝐶 = 2/6.8

⟹ 𝑅 = 6.8/2

= 3.4 Ω

Question 8: 

When a 12 V battery is connected across an unknown resistor, there is a current 

of 2.5 mA in the circuit. Find the value of the resistance of the resistor.

Answer :

According to Ohm’s law, V = IR

⟹ 𝑅 =𝑉/𝐼

Here, V = 12 V and I = 2.5 mA = 0.0025 A

Therefore,

𝑅 = 12/ 0.0025
= 4800 Ω
 = 4.8 kΩ

Question 9: 

A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5

Ω and 12 Ω, respectively. How much current would flow through the 12 Ω 

resistor?

Answer :

Total resistance of resistors when connected in series is given by

𝑅 = 𝑅1 + 𝑅2 + 𝑅3 + 𝑅4 + 𝑅5

⟹ 𝑅 = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

According to Ohm’s law, V = IR

⟹ 𝐼 = 𝑉 / 𝑅

= 9 / 13.4

= 0.67 𝐴

There is no current division occurring in a series circuit. So, the current through 

the 12 Ω resistor will be same as 0.67 A.

Question 10: 

How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?

Answer :

Let the total number of resistors be x.

Given that: 

Current I = 5 A and Potential Difference V = 220 V

According to Ohm’s law, V = IR

⟹ 𝑅 = 𝑉/𝐼

= 220/5

= 44 Ω

Now for x number of resistors of resistance 176 Ω, the equivalent resistance of 

the resistors connected in parallel is 44 Ω.

1/44 =1/176 +1/176 + 1/176 +1/176 +.... .𝑡𝑜 𝑥 𝑡𝑖𝑚𝑒𝑠

⟹ 1 / 44 = 𝑥 / 176

⟹ 𝑥 = 176/44 = 4

Therefore, 4 resistors of 176 Ω are required to draw the given amount of current.

Question 11: 

Show how you would connect three resistors, each of resistance 6 Ω, so that the

combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer :

(i). To get total 9 Ω resistance from three 6 Ω resistors, we should connect two 

resistors in parallel and the third resistor in series with the resultant. The 

combination is given as follows:

Total resistance in parallel is given by

1 / 𝑅12 = 1/𝑅1 + 1/𝑅2

⟹ 1/𝑅12 =1/6 + 1/6

= 2/6

= 1/3

⟹ 𝑅12 = 3 Ω

Now 𝑅12 and 6 Ω are connected in series, so the net resistance is given by

𝑅 = 𝑅12 + 6 Ω = 3 Ω + 6 Ω = 9 Ω

(ii). To get total 4 Ω resistance from three 6 Ω resistors, we should connect two 

resistors in series and the third resistor in parallel with the resultant. The 

combination is given as follows:

Total resistance in series is given by

𝑅12 = 𝑅1 + 𝑅2 = 6 Ω + 6 Ω = 12 Ω

Now 𝑅12 and 6 Ω are connected in parallel, so the net resistance is given by

1/𝑅 = 1/𝑅12 + 1/6

⟹ 1/𝑅 = 1/12 + 1/6

⟹ 1/𝑅 = 3/12
⟹ 1/𝑅 = 1/4

⟹ 𝑅 = 4 Ω

Question 12: 

Several electric bulbs designed to be used on a 220 V electric supply line, are 

rated 10 W. How many lamps can be connected in parallel with each other across 

the two wires of 220 V line if the maximum allowable current is 5 A?

Answer :

For one bulb:

Power P = 10 W and Potential difference V = 220 V

Using the relation for R, we have

𝑅 = 𝑉^2/𝑃

=(220)^2/10
= 4840 Ω

Let the total number of bulbs be x.

Given that: 

Current I = 5 A and Potential Difference V = 220 V

According to Ohm’s law, V = IR

⟹ 𝑅 = 𝑉/𝐼

=220/5

= 44 Ω

Now, for x number of bulbs of resistance 176 Ω, the equivalent resistance of the 

resistors connected in parallel is 44 Ω.

1/44 = 1/4840  + 1/4840 + 1/4840 +. .𝑡𝑜 𝑥 𝑡𝑖𝑚𝑒𝑠

⟹ 1/44 = 𝑥/4840

⟹ 𝑥 = 4840/44
⟹ 𝑥 = 110

Therefore, 110 bulbs of 4840 Ω are required to draw the given amount of current.

Question 13: 

A hot plate of an electric oven connected to a 220 V line has two resistance coils 

A and B, each of 24 Ω resistance, which may be used separately, in series, or in

parallel. What are the currents in the three cases?

Answer :

Given that: 

Potential difference V = 220 V and resistance of each coil R = 24 Ω

When the coil is used separately, the current in the coil is given by

𝐼 = 𝑉/𝑅

= 220/24
=55/6

= 9.16 𝐴

When the two coils are connected in series, the net resistance is given by

𝑅 = 𝑅1 + 𝑅2 = 24 Ω + 24 Ω = 48 Ω

Now, the current in the coil is given by

𝐼 = 𝑉/𝑅

= 220/48
= 55/12
= 4.58 𝐴

When the two coils are connected in parallel, the net resistance is given by

1/𝑅 = 1/24 + 1/24

1/𝑅 = 2/24

1/𝑅 = 1/12

⟹ 𝑅 = 12 Ω

Now, the current in the coil is given by

𝐼 = 𝑉/𝑅

= 220/12
 = 55/3

= 18.33 𝐴

Question 14: 

Compare the power used in the 2 Ω resistor in each of the following circuits:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in 

parallel with 12 Ω and 2 Ω resistors.

Answer :

Given that:

Potential difference, V = 6 V

(i). 1 Ω and 2 Ω resistors are connected in series. Therefore, equivalent resistance 

of the circuit, R = 1 + 2 = 3 Ω

According to Ohm’s law,

V = IR

⟹ 𝐼 = 𝑉/𝑅

= 6/3

= 2 𝐴

In series combination, the current in the circuit remains constant. Therefore power 

is given by

𝑃 = 𝐼^2𝑅 = (2)^2 × 2 = 8 𝑊

(ii). 1 Ω and 2 Ω resistors are connected in parallel. 

⟹ 𝐼 = 𝑉/𝑅

= 6/3

= 2 𝐴

In parallel combination, the voltage in the circuit remains constant. Therefore 

power is given by

𝑃 = 𝑉^2/𝑅

= 4^2/2

= 8 𝑊

Hence, in both the cases power remains same as 8W.

Question 15: 

Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected

in parallel to electric mains supply. What current is drawn from the line if the 

supply voltage is 220 V?

Answer :

For the lamp one:

Power P1 = 100 W and Potential difference V = 220 V

Therefore, 

𝐼1 = 𝑃1/𝑉

= 100 / 220
= 0.455 𝐴

For the lamp two:

Power P2 = 60 W and Potential difference V = 220 V

Therefore, 

𝐼2 = 𝑃2/𝑉

= 60/220
= 0.273 𝐴

So, the net current drawn from the supply is given by

𝐼 = 𝐼1 + 𝐼2
= 0.455 + 0.273
 = 0.728 𝐴

Question 16: 

Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 

minutes?

Answer :

Energy consumed by an electrical appliance is given by H = Pt

For the TV set:

Power W = 250 W and time t = 1 hour = 3600 seconds

So, energy consumed H = 250 × 3600 = 900000 J

For the toaster:

Power W = 1200 W and time t = 10 minutes = 600 seconds

So, energy consumed H = 1200 × 600 = 720000 J

Hence, TV set uses more energy than toaster.

Question 17: 

An electric heater of resistance 8 Ω draws 15 A from the service mains 2 hours.

Calculate the rate at which heat is developed in the heater.

Answer :

Heat developed in the heater is given by H = I2Rt

Where, I = 15 A, R = 8 Ω and time t = 2 hours

The rate at which heat is developed is given by 

𝐻 = 𝐼^2𝑅𝑡/𝑡

= 𝐼^2𝑅
= (15)^2 × 8
= 1800 𝐽/𝑠

Question 18: 

Explain the following.

a) Why is the tungsten used almost exclusively for filament of electric lamps?

b) Why are the conductors of electric heating devices, such as bread-toasters

and electric irons, made of an alloy rather than a pure metal?

c) Why is the series arrangement not used for domestic circuits?

d) How does the resistance of a wire vary with its area of cross-section?

e) Why are copper and aluminium wires usually employed for electricity

transmission?

Answer :

a) The melting point and resistivity of tungsten are very high. It does not burn 

readily at a high temperature. The electric lamps glow at very high 

temperatures. Hence, tungsten is mainly used as heating element of electric 

bulbs.

b) The conductors of electric heating devices such as bread toasters and 

electric irons are made of alloy because resistivity of an alloy is more than 

that of metals. It produces large amount of heat and do not burn easily.

c) There is voltage division in series circuits. Each component of a series 

circuit receives a small voltage for a large supply voltage. As a result, the 

amount of current decreases and the device becomes hot. Hence, series 

arrangement is not used in domestic circuits.

d) Resistance (R) of a wire is inversely proportional to its area of cross-section 

(A):

𝑅 ∝ 1/𝐴

e) Copper and aluminium wires have low resistivity. They are good 

conductors of electricity. Hence, they are usually employed for electricity 

transmission.

Dear Asif Sir